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bazaltina [42]
2 years ago
15

As electrons fall from high energy orbitals to lower orbitals, energy is released in the form of electromagnetic radiation. The

farther the electron falls, the more energy is released. Which of the following electronic transitions would produce a wave with the lowest frequency?
Chemistry
1 answer:
weeeeeb [17]2 years ago
3 0

The electronic transition that will produce the lowest frequency is an electron falling from the 3rd to the 2nd energy level.

The question is incomplete, the complete question is;

As electrons fall from high energy orbitals to lower orbitals, energy is released in the form of electromagnetic radiation. The farther the electron falls, the more energy is released. Which of the following electronic transitions would produce a wave with the lowest frequency?  

an electron falling from the 6th to the 2nd energy level  

an electron falling from the 5th to the 2nd energy level  

an electron falling from the 3rd to the 2nd energy level  

an electron jumping from the 1st to the 2nd energy level

According to Bohr's theory, energy is absorbed or emitted when an electron moves from one energy level to another. This energy often occurs as visible light of known frequency and wavelength.

The magnitude of frequency of light depends on the difference in energy between the two energy levels. If the difference between the energy levels is high, the frequency of light is also high and vice versa.

The transition from 3rd to the 2nd energy level represents a low frequency transition because the energy levels are close together.

Learn more: brainly.com/question/10675485

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8 0
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Write the chemical equations involved in this experiment and how that the rate of disappearance of [S2O8^2-] is proportional to
Ne4ueva [31]

S₂O₈²⁻

(aq) + 2I⁻

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²⁻(aq)

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(aq) + 2I⁻

(aq)

<u>Explanation:</u>

S₂O₈²⁻

(aq) + 2I⁻

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²⁻(aq)

To measure the rate of this reaction we must measure the rate of concentration change of one of  the reactants or products. To do this, we will include (to the reacting S₂O₈

²⁻  and I⁻

i) a small amount of sodium thiosulfate, Na₂S₂O₃,

ii) some starch indicator.

The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂  as soon as it is formed.

2S₂O₃²⁻

(aq) + I₂(aq) → S₄O₆²⁻

(aq) + 2I⁻

(aq)

This reaction is much faster than the previous, so the conversion of I2 back to I⁻  is  essentially instantaneous.

rate = \frac{dI2}{dt} = \frac{1/2 [S2O3^2^-]}{t}

5 0
3 years ago
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Thus, we can conclude that out of the given options, the statement even at low concentrations, a strong base is strong best relates the strength and concentration of a base.


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